Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
template <class T>
T &chmin(T &a, const T &b) {
  return a = min(a, b);
}
template <class T>
T &chmax(T &a, const T &b) {
  return a = max(a, b);
}
template <class T>
istream &operator>>(istream &is, vector<T> &v) {
  for (auto &i : v) is >> i;
  return is;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  const string delimiter = "\n";
  for (int i = (int)(0); i < (int)(v.size()); i++) {
    os << v[i];
    if (i != v.size() - 1) os << delimiter;
  }
  return os;
}
int main() {
  cin.sync_with_stdio(false);
  cout << fixed << setprecision(10);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > v(n, vector<int>(m));
  cin >> v;
  int k = *min_element(v[0].begin(), v[0].end());
  vector<pair<int, int> > anss = {pair<int, int>(-1, -1)};
  for (int i = (int)(0); i < (int)(k + 1); i++) {
    vector<vector<int> > t(n, vector<int>(m));
    t[0] = vector<int>(m, i);
    vector<pair<int, int> > ans(i, pair<int, int>(0, 0));
    for (int j = (int)(0); j < (int)(m); j++) {
      int x = v[0][j] - t[0][j];
      for (int l = (int)(0); l < (int)(x); l++) {
        ans.emplace_back(1, j);
      }
      for (int l = (int)(0); l < (int)(n); l++) {
        t[l][j] += x;
      }
    }
    bool f = true;
    for (int j = (int)(1); j < (int)(n); j++) {
      if (t[j][0] > v[j][0]) {
        f = false;
        break;
      }
      int x = v[j][0] - t[j][0];
      for (int l = (int)(0); l < (int)(x); l++) {
        ans.emplace_back(0, j);
      }
      for (int l = (int)(0); l < (int)(m); l++) {
        t[j][l] += x;
      }
    }
    if (t != v) f = false;
    if (!f) continue;
    if (anss == vector<pair<int, int> >({pair<int, int>(-1, -1)}) ||
        anss.size() > ans.size())
      anss = ans;
  }
  if (anss == vector<pair<int, int> >({pair<int, int>(-1, -1)})) {
    cout << -1 << endl;
  } else {
    cout << anss.size() << endl;
    for (auto p : anss) {
      cout << (p.first == 0 ? "row " : "col ") << p.second + 1 << endl;
    }
  }
  return 0;
}